Selection of Terms in an Arithmetic Progression

Sometimes we need to assume certain number of terms in Arithmetic Progression. The following ways are generally used for the selection of terms in an arithmetic progression.

(i) If the sum of three terms in Arithmetic Progression be given, assume the numbers as a - d, a and a + d. Here common difference is d.

(ii) If the sum of four terms in Arithmetic Progression be given, assume the numbers as a - 3d, a - d, a + d and a + 3d.

(iii) If the sum of five terms in Arithmetic Progression be given, assume the numbers as a - 2d, a - d, a, a + d and a + 2d. Here common difference is 2d.

(iv) If the sum of six terms in Arithmetic Progression be given, assume the numbers as a - 5d, a - 3d, a - d, a + d, a + 3d and a + 5d. Here common difference is 2d.

Note: From the above explanation we understand that in case of an odd number of terms, the middle term is ‘a’ and the common difference is ‘d’.

Again, in case of an even number of terms the middle terms are a - d, a + d and the common difference is 2d.


Solved examples to observe how to use the selection of terms in an arithmetic progression

1. The sum of three numbers in Arithmetic Progression is 12 and the sum of their square is 56. Find the numbers.

Solution:

Let us assume that the three numbers in Arithmetic Progression be a - d, a and a + d.

According to the problem,

Sum = 12                             and

⇒ a - d + a + a + d = 12

⇒ 3a = 12

⇒ a = 4

Sum of the squares = 56

(a - d)\(^{2}\) + a\(^{2}\) + (a + d)\(^{2}\) = 56

⇒ a\(^{2}\) - 2ad + d\(^{2}\) + a\(^{2}\) + a\(^{2}\) + 2ad + d\(^{2}\) = 56

⇒ 3a\(^{2}\) + 2d\(^{2}\) = 56

⇒ 3 × (4)\(^{2}\) + 2d\(^{2}\) = 56

⇒ 3 × 16 + 2d\(^{2}\) = 56

⇒ 48 + 2d\(^{2}\) = 56

⇒ 2d\(^{2}\) = 56 - 48

⇒ 2d\(^{2}\) = 8

⇒ d\(^{2}\) = 4

⇒ d = ± 2

If d = 3, the numbers are 4 – 2, 4, 4 + 2 i.e., 2, 4, 6

If d = -3, the numbers are 4 + 2, 4, 4 - 2 i.e., 6, 4, 2

Therefore, the required numbers are 2, 4, 6 or 6, 4, 2.

2. The sum of four numbers in Arithmetic Progression is 20 and the sum of their square is 120. Find the numbers.

Solution: 

Let us assume that the four numbers in Arithmetic Progression be a - 3d, a - d, a + d and a + 3d.

According to the problem,

Sum = 20

⇒ a - 3d + a - d + a + d + a + 3d = 20

⇒ 4a = 20

⇒ a = 5

and

Sum of the squares = 120

⇒ (a - 3d)\(^{2}\) + (a - d)\(^{2}\) + (a + d)\(^{2}\) + (a + 3d)\(^{2}\) = 120

⇒ a\(^{2}\) - 6ad + 9d\(^{2}\) + a\(^{2}\) - 2ad + d\(^{2}\) + a\(^{2}\) + 2ad + d\(^{2}\) + a\(^{2}\) + 6ad + 9d\(^{2}\) = 120

⇒ 4a\(^{2}\) + 20d\(^{2}\) = 120

⇒ 4 × (5)\(^{2}\) + 20d\(^{2}\) = 120

⇒ 4 × 25 + 20d\(^{2}\) = 120

⇒ 100 + 20d\(^{2}\) = 120

⇒ 20d\(^{2}\) = 120 - 100

20d\(^{2}\) = 20

⇒ d\(^{2}\) = 1

⇒ d = ± 1

If d = 1, the numbers are 5 - 3, 5 - 1, 5 + 1, 5 + 3 i.e., 2, 4, 6, 8

If d = -1, the numbers are 5 + 3, 5 + 1, 5 - 1, 5 - 3 i.e., 8, 6, 4, 2

Therefore, the required numbers are 2, 4, 6, 8 or 8, 6, 4, 2.

 

3. The sum of three numbers in Arithmetic Progression is -3 and their product is 8. Find the numbers.

Solution:

Let us assume that the three numbers in Arithmetic Progression be a - d, a and a + d.

According to the problem,

Sum = -3                                  and

⇒ a - d + a + a + d = -3

⇒ 3a = -3

⇒ a = -1

Product = 8

⇒ (a - d) (a) (a + d) = 8

⇒ (-1)[(-1)\(^{2}\) - d\(^{2}\)] = 8

⇒ -1(1 - d\(^{2}\)) = 8

⇒ -1 + d\(^{2}\) = 8

⇒ d\(^{2}\) = 8 + 1

⇒ d\(^{2}\) = 9

⇒ d = ± 3

If d = 3, the numbers are -1 - 3, -1, -1 + 3 i.e., -4, -1, 2

If d = -3, the numbers are -1 + 3, -1, -1 - 3 i.e., 2, -1, -4

Therefore, the required numbers are -4, -1, 2 or 2, -1, -4.

Arithmetic Progression





11 and 12 Grade Math 

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